So we see that it's a geometric progression (constant quotient) and it's going on infinitely. So we can apply the formula: $\sum_{n=0}^\infty a_1 q^n=\frac{a_1}{1-q}$ and we get: $\frac{x}{1-x^2}=-\frac{2}{3}$, but as it's only valid for an absolute value of the quotient less than one, we have to presume $|x^2|<1$. Solving the equation, we get $x=2$ or $x=-\frac{1}{2}$. The first is contrary to our assumption so $x=-\frac{1}{2}$.